Calculus, find the limit, Exp vs Power? $\lim_{x\to\infty} \frac{e^x}{x^n}$
n is any natural number. 
Using L'hopital doesn't make much sense to me. I did find this in the book: 
"In a struggle between a power and an exp, the exp wins."
Can I refer that line as an answer? If the fraction would have been flipped, then the limit would be zero. But in this case I the limit is actually $\infty$
 A: Repeated use of L'Hôpital's rule ($n$ times):
$\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{n}}=\underset{%
x\rightarrow \infty }{\lim }\dfrac{e^{x}}{nx^{n-1}}=\underset{x\rightarrow
\infty }{\lim }\dfrac{e^{x}}{n(n-1)x^{n-2}}=\cdots =$ $=\cdots=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{n(n-1)\cdots
3\cdot x^{2}}=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{%
n(n-1)\cdots 3\cdot 2x}=$ $=\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{%
n(n-1)\cdots 3\cdot 2\cdot 1}=\underset{x\rightarrow \infty }{\lim }\dfrac{%
e^{x}}{n!}=\infty$

Added: 
To convince yourself: If you had $\underset{x\rightarrow \infty }{\lim }\dfrac{e^{x}}{x^{10}}$ you would have to apply L'Hôpital's rule ten times.
Added 2: Plot of $\dfrac{e^{x}}{x^{3}}$

A: Use the fact that $e^x \ge \left( 1 + \frac{x}{n} \right)^n$ for any $n > 0$.
A: Why isn't L'Hôpital a good solution? Just use induction.
A: HINT: One way of looking at this would be: $$\frac{1}{x^{n}} \biggl[ \biggl(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}\biggr) + \frac{x^{n+1}}{(n+1)!} + \cdots \biggr]$$
I hope you understand why i put the brackets inside those terms.
A: I believe the problem is tailor-made for repeated application of L'Hopital's Rule, but here are some thoughts ...
You could note that $e^{x} = (e^{x/n})^n$, and consider $\left( \lim \frac{e^{x/n}}{x}\right)^n$, so that you are comparing an exponential to a single power of $x$, which might be a bit less daunting for you.
A bit more cleanly, and to make the numerator and denominator match better, define $y := \frac{x}{n}$. Then $$\frac{e^{x}}{x^n}=\frac{e^{ny}}{(ny)^n}=\frac{\left(e^{y}\right)^n}{n^n y^n}=\frac{1}{n^n}\frac{\left(e^{y}\right)^n}{y^{n}}=\frac{1}{n^n}\left(\frac{e^y}{y}\right)^n$$
Since $n$ is a constant, you can direct your limiting attention to $\frac{e^y}{y}$ (as $y \to \infty$, of course).
A: Let's consider the limit to infinity when taking log of that expression:
$$\lim_{x\rightarrow\infty} \ln{\frac{e^x}{x^n}} = \lim_{x\rightarrow\infty}x-n\ln(x)=\lim_{x\rightarrow\infty}x(1-\frac{n\ln(x)}{x})=\infty$$
Therefore, the limit is $\infty$.
The proof is complete.
